Question: Simplify and expand the following expression: $ \dfrac{4}{3x + 15}+ \dfrac{2}{5x - 40}+ \dfrac{2}{x^2 - 3x - 40} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{4}{3x + 15} = \dfrac{4}{3(x + 5)}$ We can factor a $5$ out of denominator in the second term: $ \dfrac{2}{5x - 40} = \dfrac{2}{5(x - 8)}$ We can factor the quadratic in the third term: $ \dfrac{2}{x^2 - 3x - 40} = \dfrac{2}{(x + 5)(x - 8)}$ Now we have: $ \dfrac{4}{3(x + 5)}+ \dfrac{2}{5(x - 8)}+ \dfrac{2}{(x + 5)(x - 8)} $ The least common multiple of the denominators is: $ 15(x + 5)(x - 8)$ In order to get the first term over $15(x + 5)(x - 8)$ , multiply by $\dfrac{5(x - 8)}{5(x - 8)}$ $ \dfrac{4}{3(x + 5)} \times \dfrac{5(x - 8)}{5(x - 8)} = \dfrac{20(x - 8)}{15(x + 5)(x - 8)} $ In order to get the second term over $15(x + 5)(x - 8)$ , multiply by $\dfrac{3(x + 5)}{3(x + 5)}$ $ \dfrac{2}{5(x - 8)} \times \dfrac{3(x + 5)}{3(x + 5)} = \dfrac{6(x + 5)}{15(x + 5)(x - 8)} $ In order to get the third term over $15(x + 5)(x - 8)$ , multiply by $\dfrac{15}{15}$ $ \dfrac{2}{(x + 5)(x - 8)} \times \dfrac{15}{15} = \dfrac{30}{15(x + 5)(x - 8)} $ Now we have: $ \dfrac{20(x - 8)}{15(x + 5)(x - 8)} + \dfrac{6(x + 5)}{15(x + 5)(x - 8)} + \dfrac{30}{15(x + 5)(x - 8)} $ $ = \dfrac{ 20(x - 8) + 6(x + 5) + 30} {15(x + 5)(x - 8)} $ Expand: $ = \dfrac{20x - 160 + 6x + 30 + 30}{15x^2 - 45x - 600} $ $ = \dfrac{26x - 100}{15x^2 - 45x - 600}$